Question: The first three stages of a pattern are shown below, in which each line segment represents a toothpick. If the pattern continues such that at each successive stage, three toothpicks are added to the previous arrangement, how many toothpicks are necessary to create the arrangement for the 250th stage? [asy] size(150); defaultpen(linewidth(0.7)); void drawSquare(pair A){ draw((A.x + 0.1,A.y)--(A.x + 0.9,A.y)); draw((A.x,A.y + 0.1)--(A.x,A.y + 0.9)); draw((A.x + 1,A.y + 0.1)--(A.x + 1,A.y + 0.9)); draw((A.x + 0.1,A.y + 1)--(A.x + 0.9,A.y + 1)); } int k = 0; for(int i = 1; i <= 3; ++i){ for(int j = 0; j < i; ++j){ drawSquare((k,0)); ++k; } draw((k+0.1,0.5)--(k+0.9,0.5),EndArrow); ++k; } label("$\cdots$",(k,0.5)); [/asy]
Solution: The number of toothpicks in each stage form an arithmetic sequence. The first term in this arithmetic sequence is 4, and the common difference is 3 (the number of toothpicks added to get to the next stage), so the number of toothpicks used in the 250th stage is $4 + 3 \cdot 249 = \boxed{751}$.